Need to step down a lead acid battery from 12vdc to 10vdc 350mA

i would like to run a 6pcs 1W white led lights from a lead acid battery. A Single bulb input voltage is 3.2-3.6V & 350ma. I connected every 3 led's in Series (2 Groups) . Can I use 7810 IC or how can I do it. Sir Please send me a diagram . lead acid are in full-charge 14.5v. , in normal 12v, low 10v

Did you make this instructable?

Find a low dropout regulator (the output voltage does not matter, but lower = better), and wire it up with a resistor between the output pin and the ground pin. attach the ground pin to the positive of an LED, and the negative to ground.

The regulator will maintain a constant voltage across that shunt resistor, and in order to do that, a constant current must flow though the LEDs. So just use ohms law to calculate the resistance needed for your regulator's output voltage.

Say, you have a 3V regulator. To have a 350mA output, the equation is V =IR, 3=.35A*R, R = 8.57 ohms. But since the voltage drop is over 3V (resistor + regulator) the input voltage has to be higher. a LM317T is essentially just a 1.25V voltage regulator, so that would work best in your situation.

A better solution would be to use a buck-boost converter with a current-limited (not protected) output.

You will need 3 regulators for 6 LEDs. You can mount the regulators on a common heatsink but they must be electrically insulated from the heatsink as the metal tab is connected to pin2, Vout. Also important if you are mounting the heatsink inside a car as you could short out the regulators.

You /can/ do it with a 7810 but that would present some other problems. The LM7810 has a dropout of about 2.5v so it'd need a minimum of 12.5v to work correctly.

To solve the dropout problem, you could use a lower voltage regulator and have more in parallel such as using an LM7806 with two in series and three parallel. Bear in mind that the standard TO-220 model LM78xx voltage regulators typically only do up to 1.5A so any more than three in parallel and you'd probably have to use the larger TO-3 packaged ones to cope with the higher current. There are also other types of voltage regulators with lower dropouts you could use, but they tend to do less current. You tend to get either high dropout and high current or low dropout and low current but not both, you'd have to do some research to see if there's one suitable.

You also said they are 1W LEDs with voltage 3.2V and 350mA current, but that doesn't quite add up. If the current is correct then 1/0.35 is ~2.9V or if the voltage is correct then they should be 310mA. 350mA at 3.2V would be 1.1W. It might seem like an insignificant difference, but you could burn out the LEDs or shorten the life of them if it's incorrect. 2.9v seems more likely but LEDs tend to range between about 2.1-3.3v.

You could also do as mpilchfamily said and have 4 in series and then 2 directly connected to the 12v with different resistors (a calculator suggests 2.4 and 20 ohm) or even create a voltage divider, but I would not recommend it for something this high powered as it'll drain the battery in no time and you'll need seriously beefy resistors for that.

I'd just suggest looking at online calculators such as this to calculate the resistors required. You also probably won't get away with standard 1/4w resistors so you'll probably need 0.6w, 1w or even 2w resistors.

Even with a regulator, you should probably use resistors as well to avoid damage to the LEDs, use the same calculator for that with the input voltage being the output from the regulator.